2005 Alabama ARML TST Problems/Problem 13

Problem

There is one natural number with exactly 6 positive divisors, the sum of whose reciprocals is 2. Find that natural number.

Solution

Let the number be , and let and be primes. Therefore, one of the following is true:

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• 

For the first one, the sum of the reciprocals of the divisors of is therefore . The smallest prime (2) makes that less than 2, and if gets bigger, then that expression gets smaller, so there is absolutely no way that . So the second case is true.

$\begin{eqnarray}1+\dfrac{1}{p_1}+\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_2^2}+\dfrac{1}{p_2}+\dfrac{1}{p_2^2}=2\\

\dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1}{p_1p_2^2}=\dfrac{(p^2+p_2+1)(p_1+1)}{p_1p_2^2}=2\\ p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\ p_1(p_2^2-p_2-1)=p_2^2+p_2+1

\end{eqnarray}$ (Error compiling LaTeX. Unknown error_msg)

Therefore, . Now the only way that that is possible is when . Solving for , we get that . Checking, the sum of the reciprocals of the divisors of indeed sum to 2, and 28 does have 6 factors.

See also

2005 Alabama ARML TST (Problems)
Preceded by: Problem 11	Followed by: Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15